Question

In the given figure, ABCD is a ||gm and E is the midpoint of BC. Also, DE and All when produced meet at F. Then,

(a) $AF=\frac{3}{2}AB$

(b) AF = 2AB

(c) AF = 3AB

(d) $A{F}^2=2A{B}^2$

Answer 1

ANSWER:
(b) AF = 2 AB
Explanation:
I​ n parallelogram ABCD, we have:
AB || DC
∠DCE = ∠​ EBF
(Alternate interior angles)In ∆ DCE and ​ ∆ BFE, we have:
∠DCE = ∠ EBF
(Proved above)
∠DEC = ∠ BEF
(Vertically opposite angles)
BE = CE
( Given)
i.e., ∆ DCE ≅​ ∆ BFE (By ASA congruence rule)
∴ DC = BF
(CPCT)
But DC= AB, as ABCD is a parallelogram.
∴ DC = AB = BF
...(i)
Now, AF = AB + BF
From (i), we get:
∴ AF = AB + AB = 2AB